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Those Darn CIRCUITS!

  1. Sep 22, 2009 #1
    !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

    Forgive me for taking your time for such a minor mistake, but I was thinking the battery was a 10V battery, but it's 12.

    But you did help me see that, thanks.
     
    Last edited: Sep 22, 2009
  2. jcsd
  3. Sep 22, 2009 #2

    rl.bhat

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    Even though the values of capacitors are given, you have not used them in your solution.
    Do it in step up by step.
    What is Ceq. of C1 and C3? What is Ceq. of C2 and C4?
    What is Ceq. of C1,C2,C3 and C4?
    What is the total charge drawn from the battery? How this charge is distributed between the two branches?
     
  4. Sep 22, 2009 #3
    I found the capacitance of C13, and I wrote out the equation I used in my first post. Then, to find the voltage across C13, I used the idea that the voltage across capacitors in parallel are equal and are equal to their equivalent capacitor. C13 and C24 are in parallel, and their equivalent capacitor is C1234. C1234 obviously has a voltage across it equal to the battery. So C13 has voltage across it equal to the battery, a value given in the problem, so I know the voltage across C13.

    So, knowing the voltage and capacitance, I multiplied the numbers and came up with 7.5 micro coulombs.

    But again, my number isn't right. If my reasoning is correct, I just can't find my calculation error. But I think that my reasoning isn't correct, but by everything that I know, I can't see where.
     
  5. Sep 22, 2009 #4

    rl.bhat

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    What is C13?
    How did you get 7.5 μC?
     
  6. Sep 22, 2009 #5
    Well since C1 and C3 are in series, their total capacitance would be .75 microF. C2 and C4 would have capacitance .875, thus total capacitance for the system would be 1.625.

    You have a 12V battery, times 1.625 total capacitance = 19.5 microcoulombs for the system (charge).

    Now, voltage on the wires is a little confusing since I'm not sure if you calculate it for each wire in series, or the whole parallel and series combination (which would give you 1.625, then dividing 19.5/1.625 would be 12, which you would then use as the voltage across each capacitor in your Q=CV equation with the C of each individual capacitor).

    However, I have a feeling each wire has a different voltage, in which case the upper wire would be .75 and bottom .875. Then, 19.5/.75= 26V across capacitors in top wire and 19.5/.875=22.28V for the bottom wire, which you would then use in your Q=CV, using the appropriate voltage for the right capacitor.

    BUT, I am not sure which scenario is the case and hopefully someone can clarify that as would help clarify the concept for me as well.
     
  7. Sep 22, 2009 #6
    C13 is

    (1 / 3) + (1 / 1) = 1 / C13
    C13 = 0.75 micro F

    And then since I think the voltage across C13 is 10 volts, I multiply 0.00000075C by 10V (b/c q = CV) and get 7.5 micro coulombs.
     
  8. Sep 22, 2009 #7
    !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

    Forgive me for taking your time for such a minor mistake, but I was thinking the battery was a 10V battery, but it's 12.

    But you did help me see that, thanks.
     
  9. Sep 22, 2009 #8

    rl.bhat

    User Avatar
    Homework Helper

    Your calculation is correct. Only change the voltage.
     
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